Question: Let $f(x)=3x^5-2x^4-4x^3$. $f'(x)=$
Solution: According to the sum rule, the derivative of $3x^5-2x^4-4x^3$ is the sum of the derivatives of $3x^5$, $-2x^4$, and $-4x^3$. The derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ For example, this is the derivative of the first term: $\begin{aligned}\dfrac{d}{dx}(3x^5)&=3\dfrac{d}{dx}(x^5)&&\gray{\text{Constant multiple rule}}\\\\ &=3\cdot (5x^4)&&\gray{\text{Power rule}}\\ \\ &=15x^4\end{aligned}$ Here is the complete differentiation process: $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}(3x^5-2x^4-4x^3) \\\\ &=3\dfrac{d}{dx}(x^5)-2\dfrac{d}{dx}(x^4)-4\dfrac{d}{dx}(x^3)&&\gray{\text{Basic differentiation rules}} \\\\ &=3\cdot 5x^4-2\cdot4x^3-4\cdot 3x^2&&\gray{\text{The power rule}} \\\\ &=15x^4-8x^3-12x^2 \end{aligned}$ In conclusion, $f'(x)=15x^4-8x^3-12x^2$.